How many grams of AgNo3 are required to make 25 mL of a .80M solution?
M = moles/L
Substitute L and M, solve for moles.
moles = grams/molar mass
Substitute moles and molar mass and solve for grams.
To calculate the number of grams of AgNO3 required to make a 25 mL solution with a concentration of 0.80 M, you need to use the equation:
Molarity (M) = Moles (mol) / Volume (L)
First, convert the volume from mL to L:
25 mL = 25 / 1000 = 0.025 L
Next, rearrange the equation to solve for moles:
Moles (mol) = Molarity (M) * Volume (L)
Moles (mol) = 0.80 * 0.025
Moles (mol) = 0.020
The molar mass of AgNO3 can be found in the periodic table:
Ag: 107.87 g/mol
N: 14.01 g/mol
O: 16.00 g/mol (three oxygen atoms in AgNO3)
Molar mass (AgNO3) = 107.87 + 14.01 + (3 * 16.00) = 169.87 g/mol
Finally, calculate the mass using the equation:
Mass (g) = Moles (mol) * Molar mass (g/mol)
Mass (g) = 0.020 * 169.87
Mass (g) = 3.3974 grams
Therefore, you would need approximately 3.40 grams of AgNO3 to make a 25 mL solution with a concentration of 0.80 M.
To calculate the number of grams of AgNo3 required to make a 25 mL solution with a concentration of 0.80 M, you need to use the formula:
Molarity (M) = (Number of moles) / (Volume in liters)
First, convert the given volume from milliliters to liters:
25 mL = 0.025 L
Next, rearrange the formula to solve for the number of moles:
(Number of moles) = Molarity (M) * Volume (L)
Plugging in the given values:
(Number of moles) = 0.80 mol/L * 0.025 L
By multiplying the molarity and volume together, you find that there are 0.02 moles of AgNo3 required to make a 25 mL solution.
To convert moles to grams, you need to use the molar mass of AgNo3, which is 169.87 g/mol.
(Number of grams) = (Number of moles) * (Molar mass)
Plugging in the values:
(Number of grams) = 0.02 mol * 169.87 g/mol
By multiplying the number of moles by the molar mass, you find that you need approximately 3.4 grams of AgNo3 to make a 25 mL solution with a concentration of 0.80 M.